What is the extreme f(x)=x^2+y^2-8x+y ?

The extreme f(x)=x^2+y^2-8x+y is Minimum(4,-1/2)

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extreme f(x)=x^2+y^2-8x+y

extreme points

f (x) = x^{2} + y^{2} - 8 x + y

Minimum (4, - \frac{1}{2})

Solution steps

Find the critical points:

(4, - \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(4, - \frac{1}{2}) :

Minimum

Minimum (4, - \frac{1}{2})

f (x) = 12 x^{3} - 9 x^{2} - 108 x + 4

f (x) = 3 x^{2} + 6 x - 6

f (x) = 3 x - 3

f (x) = - x^{3} - x^{2} + 5 x + 1

\frac{30}{x} + 3 π x^{2}

What is the extreme f(x)=x^2+y^2-8x+y ?
The extreme f(x)=x^2+y^2-8x+y is Minimum(4,-1/2)